Integrand size = 10, antiderivative size = 111 \[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}-\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2} \]
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Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6667, 6874, 3378, 3384, 3379, 3382} \[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=-\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Shi}(a+b x)}{2 x^2}-\frac {b \sinh (a+b x)}{2 a x} \]
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Rule 3378
Rule 3379
Rule 3382
Rule 3384
Rule 6667
Rule 6874
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {1}{2} b \int \frac {\sinh (a+b x)}{x^2 (a+b x)} \, dx \\ & = -\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {1}{2} b \int \left (\frac {\sinh (a+b x)}{a x^2}-\frac {b \sinh (a+b x)}{a^2 x}+\frac {b^2 \sinh (a+b x)}{a^2 (a+b x)}\right ) \, dx \\ & = -\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {b \int \frac {\sinh (a+b x)}{x^2} \, dx}{2 a}-\frac {b^2 \int \frac {\sinh (a+b x)}{x} \, dx}{2 a^2}+\frac {b^3 \int \frac {\sinh (a+b x)}{a+b x} \, dx}{2 a^2} \\ & = -\frac {b \sinh (a+b x)}{2 a x}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {b^2 \int \frac {\cosh (a+b x)}{x} \, dx}{2 a}-\frac {\left (b^2 \cosh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx}{2 a^2}-\frac {\left (b^2 \sinh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx}{2 a^2} \\ & = -\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {\left (b^2 \cosh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx}{2 a}+\frac {\left (b^2 \sinh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx}{2 a} \\ & = \frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}-\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\frac {b^2 x^2 \text {Chi}(b x) (a \cosh (a)-\sinh (a))-a b x \sinh (a+b x)+b^2 x^2 (-\cosh (a)+a \sinh (a)) \text {Shi}(b x)-a^2 \text {Shi}(a+b x)+b^2 x^2 \text {Shi}(a+b x)}{2 a^2 x^2} \]
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\[\int \frac {\operatorname {Shi}\left (b x +a \right )}{x^{3}}d x\]
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\[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x + a\right )}{x^{3}} \,d x } \]
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\[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\int \frac {\operatorname {Shi}{\left (a + b x \right )}}{x^{3}}\, dx \]
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\[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x + a\right )}{x^{3}} \,d x } \]
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\[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x + a\right )}{x^{3}} \,d x } \]
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Timed out. \[ \int \frac {\text {Shi}(a+b x)}{x^3} \, dx=\int \frac {\mathrm {sinhint}\left (a+b\,x\right )}{x^3} \,d x \]
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